Whispers from Old Graves
Whispers from Old Graves
This word BOTE-SWAINE deserves notice, not only because it is the first word of dialogue in a book of about 900,000 words, but because of the mathematical probability of the occurrence of these 10 letters while using a 24 letter Elizabethan alphabet.
According to Derek Rowntree ("Probability without Tears" 1984, Barnes & Noble) where n = number of letters in the alphabet and r = number of letters in BOTESWAINE, the formula is:n! / ((n-r)! x r!)
The arithmetic is:
24! / ((24-10)! x 10!) = 1,961,256Because there are 900,000+ words in the First Folio, this number 1,961,156 must be multiplied by 900,000 to find the odds against one. The product is 1,965,130,400,000.
Call it 1 trillion, 765 billion, 130 million to one. William F. Friedman once wrote that billions were enough for him.
As an alternative to this method of calculating probability, the following is equally applicable:
Shakespeare had a vocabulary of 15,000+ words according to most authorities. From these he could choose one. His 1623 Folio of the plays contains 900,000+ words. He could put his chosen word anywhere.
The odds against putting the word "Bote-swaine" anywhere are 15,000 times 900,000 = 13,500,000,000. His choice was: as the first word of dialogue.
To explain further, the probability of selecting a particular word from the author's vocabulary is one in 15,000. The probability of inserting that word in a particular place in the First Folio is one in 900,00.
Again, according to Derek Rowntree our expert in this science, the combined probability is the product of the separate, individual probabilities. This is called the multiplication rule. It is one that dismays all Stratfordians.
Why do we have two calculations? It is a matter of definition. If Bote-swaine is not a word, the first method applies, because all possible combinations of the 24 letter alphabet are presumed. The selection is random.
If Bote-swaine is a word, then it must be assumed to be part of the First Folio vocabulary, and the second method is valid.
There are arguments for both suppositions, but the point need not be determined. We have an appropriate calculation to satisfy either assumption, one computing the odds at trillions and the other at more than 13 billions to one.
"The mathematical theory of probability can be applied, and the chances calculated exactly. If the cryptanalyst finds a certain key and (on the basis of the way it is built up) he calculates that the chances of its appearing by accident are one in one thousand million his confidence in the solution will be more than justified (The Shakespearean Ciphers Examined, William F. Friedman)
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